A 1.00-cm high object is placed 3.00 cm to the left of a converging lens of foca

Question

A 1.00-cm high object is placed 3.00 cm to the left of a converging lens of focal length 8.00 cm. A diverging with a focal length of magnitude 16.00 cm is 6.00 cm to the right of the converging lens.

(a) Draw a ray diagram for this situation and locate the position of the final image. Your diagram must contain all three principal rays for both lenses.

(b) Calculate the position of the final image, relative to the diverging lens.

(c) Calculate the height of the final image.

(d) Is the final image inverted or upright relative to the original object?

(e) Is the final image real or virtual?

Explanation / Answer

y1 = 1cm , u1 = 3 cm, f = 8 cm

From lens formula:

for converging lens

(1/f) =(1/u)+(1/v)

(1/v1) = (1/8) - (1/3)

v 1=-4.8 cm (in front of converging lens)

for the diverging lens :: f = -16 cm
u2 = 4.8 + 6 = 10.8 cm

(1/f) =(1/u)+(1/v)

(1/v2) = (1/-16) - (1/10.8)

v2 =-6.45 cm (in front of diverging lens)

Position of final image is 6.45 cm infront of diverging lens

(b) the height of the final image:

y2/y1 = -(v1/u1) * (v2/u2)

y2/1 = (4.8/3)*(6.45/10.8)

y2 = 0.96 cn

(d) lateral magnification is positive.

so final image is upright

(e) image distance is negative so final image is virtual

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